业务规则:仅当trans_code分组的总数(trans_amount)>时,才获取客户的总cust_points数。 0.
对于客户#1,date_code级别(代码10)的摘要是> 0所以cust_points总数= 70。
对于客户#2,仅代码20组总数> 0总共只有75个cust_points
这是我的疑问:
with customers as
(select '1' as cust_id, 10 as date_code, 30 as cust_points from dual union all
select '1' as cust_id, 10 as date_code, 40 as cust_points from dual union all
select '1' as cust_id, 20 as date_code, 22 as cust_points from dual union all --These points should not total because trans_amount sum for code 20 is less than 0
select '1' as cust_id, 40 as date_code, 33 as cust_points from dual union all -- These points should not total because there is not trans_amounts > 0 for date_code
select '2' as cust_id, 10 as date_code, 20 as cust_points from dual union all
select '2' as cust_id, 20 as date_code, 65 as cust_points from dual union all
select '2' as cust_id, 20 as date_code, 10 as cust_points from dual
),
transactions_row as
(
select '1' as cust_id, '10' as trans_code, 10.00 as trans_amount from dual union all
select '1' as cust_id, '20' as trans_code, -15.00 as trans_amount from dual union all
select '1' as cust_id, '20' as trans_code, -20.00 as trans_amount from dual union all
select '1' as cust_id, '20' as trans_code, -10.00 as trans_amount from dual union all
select '1' as cust_id, '30' as trans_code, 30.00 as trans_amount from dual union all
select '1' as cust_id, '20' as trans_code, -20.00 as trans_amount from dual union all
select '2' as cust_id, '10' as trans_code, -50.00 as trans_amount from dual union all
select '2' as cust_id, '20' as trans_code, 20.00 as trans_amount from dual
)
select cust_id,
sum(cust_points)
from customers
where cust_id in
(
select cust_id
from (
select cust_id, trans_code, sum(trans_amount)
from transactions_row
group by cust_id, trans_code
having sum(trans_amount) > 0
)
)
group by cust_id
Desired Results
CUST_ID CUST_POINTS
1 70 /* (30 because total trans_amount for tran_code(10) > 0 +
40 because total trans_amount for tran_code(10) > 0) */
2 75 /* Do not include the 20 points because total trans_amt for 10 < 0 */
答案 0 :(得分:1)
这是使用exists
的一种方式:
select cust_id,
sum(cust_points)
from customers c
where exists (
select 1
from transactions_row tr
where tr.trans_code = c.date_code
and tr.cust_id = c.cust_id
group by tr.trans_code, tr.cust_id
having sum(tr.trans_amount) > 0
)
group by cust_id